Question: A circle with radius $6$ has a sector with a $40^\circ$ central angle. What is the area of the sector? ${36\pi}$ $\color{#9D38BD}{40^\circ}$ ${4\pi}$ ${6}$
Solution: First, calculate the area of the whole circle. Then the area of the sector is some fraction of the whole circle's area. $A_c = \pi r^2$ $A_c = \pi (6)^2$ $A_c = 36\pi$ The ratio between the sector's central angle $\theta$ and $360^\circ$ is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{360^\circ} = \dfrac{A_s}{A_c}$ $\dfrac{40^\circ}{360^\circ} = \dfrac{A_s}{36\pi}$ $\dfrac{1}{9} = \dfrac{A_s}{36\pi}$ $\dfrac{1}{9} \times 36\pi = A_s$ $4\pi = A_s$